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weeeeeb [17]
3 years ago
7

Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-chlorobutane and 1-butanol by fra

ctional distillation?
A.
Both 1-chlorobutane and 1-butanol are polar.
B.
Both 1-chlorobutane and 1-butanol are nonpolar.
C.
The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol.
D.
The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol.
Chemistry
1 answer:
KatRina [158]3 years ago
6 0

Answer:

The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol

Explanation:

Fractional distillation is a separation process based on difference in boiling point of two compounds.

1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence  dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.

1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.

Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.

So, boiling point of 1-butanol is much higher than 1-chlorobutane.

Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.

So, option (D) is correct.

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A sample of nitrogen gas had a volume of 500. ml, a pressure in its closed container of 740 torr and a temperature 25 degrees c
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Question:
              <span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?

Answer:

Data Given:
                   V</span>₁  =  500 mL

                   P₁  =  740 torr

                   T₁  =  25 °C + 273  =  298 K

                   V₂  =  ?

                   P₂  =  760 torr

                   T₂  =  50 °C + 273  =  323 K

Solution:
             Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,

                      P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                      V₂  =  (P₁ V₁ T₂) ÷ (T₁ P₂)

Putting Values,
  
                      V₂  =  (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)

                      V₂  =  527.68 mL
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